1. Tardigrade
  2. Question
  3. Chemistry
  4. The following concentrations were obtained for the formation of NH 3(g) from N 2(g) and H 2(g) at equilibrium and at 500 K :[ N 2]=1 × 10-2 M,[ H 2]=2 × 10-2 M and [ NH 3]=2 × 10-2 M. The equilibrium constant, Kc, for the reaction N 2(g)+3 H 2(g) leftharpoons 2 NH 3(g) at 500 K is

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Q. The following concentrations were obtained for the formation of $NH _3(g)$ from $N _2(g)$ and $H _2(g)$ at equilibrium and at
$500 K :\left[ N _2\right]=1 \times 10^{-2} M,\left[ H _2\right]=2 \times 10^{-2} M$ and $\left[ NH _3\right]=2 \times 10^{-2} M$.
The equilibrium constant, $K_c$, for the reaction
$N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g)$ at $500 K$ is

KEAMKEAM 2021

Solution:

${\left[N_2\right]=1 \times 10^{-2} M}$
$ {\left[H_2\right]=2 \times 11^{-2} M} $
$ {\left[ NH _3\right]=2 \times 10^{-2} M} $
$ N_2(g)+3 H_2(g) \rightleftharpoons 2 NH _3(g) $
$\text { Equilibrium constant, } K_c=\frac{\left[ NH _3\right]^2}{\left[ N _2\right]\left[ H _2\right]^3} $
$ \therefore K_c=\frac{\left(2 \times 10^{-2}\right)^2}{\left(1 \times 10^{-2}\right)\left(2 \times 10^{-2}\right)^3}$
$ =\frac{\left(2 \times 10^{-2}\right)^{2-3}}{1 \times 10^{-2}}=\frac{1}{\left(1 \times 10^{-2}\right)\left(2 \times 10^{-2}\right)} $
$=0.5 \times 10^4 \, mol ^{-2} \,dm ^6$
$ =5 \times 10^3 \, mol ^{-2} \,dm$